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Question Number 215550 by MrGaster last updated on 10/Jan/25

$$\mathbf{Let}{\mathit{u}}^{\left(1\right)},{\mathit{u}}^{\left(2\right)}\mathbf{s}.\mathbf{t}.\{\begin{array}{l}{\mathit{u}}_{\mathit{t}\mathit{t}}^{\left(1\right)}=(\frac{{\partial }^2}{\partial {\mathit{x}}_1^2}+\frac{{\partial }^2}{\partial {\mathit{x}}_i^2}){\mathit{u}}^{\left(1\right)}\\ {\mathit{u}}^{\left(1\right)}({\mathit{x}}_1,{\mathit{x}}_2,0)=\mathit{\psi }({\mathit{x}}_1,{\mathit{x}}_2)\\ {\mathit{u}}^{\left(1\right)}({\mathit{x}}_1,{\mathit{x}}_2,0)=0\end{array},\{\begin{array}{l}{\mathit{u}}_{\mathit{t}\mathit{t}}^{\left(2\right)}=(\frac{{\partial }^2}{\partial {\mathit{x}}_1^2}+\frac{{\partial }^2}{\partial {\mathit{x}}_2^2}+{\mathit{c}}^2){\mathit{u}}^{\left(2\right)}\\ {\mathit{u}}^{\left(2\right)}({\mathit{x}}_1{\mathit{x}}_2,0)=0\\ {\mathit{u}}_{\mathit{t}}^{\left(2\right)}({\mathit{x}}_1,{\mathit{x}}_2,0)=\mathit{\psi }({\mathit{x}}_1,{\mathit{x}}_2)\end{array}$$$$\mathrm{prove}:{\mathit{u}}^{\left(2\right)}({\mathit{x}}_1,{\mathit{x}}_2,\mathit{t})=\frac{1}{2\mathit{\pi }}\int {\int }_{{\mathit{\xi }}_1^2+{\mathit{\xi }}_2^2\le {\mathit{t}}^2}\frac{{\mathit{e}}^{{\mathit{\xi }}_2\mathit{c}}{\mathit{u}}^{\left(1\right)}({\mathit{x}}_1,{\mathit{x}}_2,{\mathit{\xi }}_1)\mathit{d}{\mathit{\xi }}_1\mathit{d}{\mathit{\xi }}_2}{\sqrt{{\mathit{t}}^2-{\mathit{\xi }}_1^2-{\mathit{\xi }}_2^2}}$$

Answered by MrGaster last updated on 03/Feb/25

$$(x,t)=\frac{1}{2\pi }\int {\int }_{{\xi }_1^2+{\xi }_2^2\le t^2}\frac{e^{\xi 2c}\varphi ({\xi }_1,{\xi }_2)d{\xi }_{1}d{\xi }_2}{\sqrt{t^2-{\xi }_1^2-{\xi }_2^2}}$$$$\mathrm{Then},u^{\left(2\right)}(x_1,x_2,t)=\frac{\partial }{\partial t}\left(tG(x,t)\right)$$$$\mathrm{By}{\mathrm{Duhamel}}^{,}\mathrm{s}\mathrm{principle},u^{\left(2\right)}(x_1,x_2,t)={\int }_0^{t}G(x_1,x_2,\tau )d\tau$$$$\mathrm{Thus},u^{\left(2\right)}(x_1,x_2,t)=\frac{1}{2\pi }{\int }_0^t\int {\int }_{{\xi }_1^2+{\xi }_2^2\le {(t-\tau )}^2}\frac{e^{{\xi }_{2}c}\psi (x_1,x_2,\tau )d{\xi }_{1}d{\xi }_2}{\sqrt{{(t-\tau )}^2-{\xi }_1^2-{\xi }_2^2}}d\tau$$$${\mathit{u}}^{\left(2\right)}({\mathit{x}}_1,{\mathit{x}}_2,\mathit{t})=\frac{1}{2\mathit{\pi }}\int {\int }_{{\mathit{\xi }}_1^2+{\mathit{\xi }}_2^2\le {\mathit{t}}^2}\frac{{\mathit{e}}^{{\mathit{\xi }}_2\mathit{c}}{\mathit{u}}^{\left(1\right)}({\mathit{x}}_1,{\mathit{x}}_2,{\mathit{\xi }}_1)\mathit{d}{\mathit{\xi }}_1\mathit{d}{\mathit{\xi }}_2}{\sqrt{{\mathit{t}}^2-{\mathit{\xi }}_1^2-{\mathit{\xi }}_2^2}}$$$$[\mathrm{Q}.\mathrm{E}.\mathrm{D}]$$

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